Given the vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} + \mathbf{k}$, find the exact area of the parallelogram formed by these two vectors.

1. Recognize that the area of a parallelogram defined by two vectors is exactly the magnitude of their vector cross product: Area $= |\mathbf{a} \times \mathbf{b}|$. 2. Write the vectors in column format for easier calculation: $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}$. 3. Compute the components of the cross product: $x = (2)(1) - (-1)(0) = 2$. 4. Compute the $y$-component (remembering the negative sign): $y = -((1)(1) - (-1)(3)) = -(1 - (-3)) = -4$. 5. Compute the $z$-component: $z = (1)(0) - (2)(3) = -6$. So, $\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 2 \\ -4 \\ -6 \end{pmatrix}$. 6. Calculate the magnitude to find the area: $\sqrt{2^2 + (-4)^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = \mathbf{2\sqrt{14}}$.

Find the acute angle between the line $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and the line $L_2: \mathbf{r} = \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix} + s \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}$. Give your answer in degrees to one decimal place.

1. Identify the direction vectors of the two lines, ignoring the position vectors: $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}$. 2. Calculate the scalar product: $\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(4) + (2)(-1) = 2 - 4 - 2 = -4$. 3. Find the magnitude of each direction vector: $|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$, and $|\mathbf{d}_2| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{18} = 3\sqrt{2}$. 4. Substitute these values into the angle formula, using the absolute value of the dot product to find the acute angle: $\cos \theta = \frac{|-4|}{3 \times 3\sqrt{2}} = \frac{4}{9\sqrt{2}}$. 5. Evaluate the inverse cosine on your GDC: $\theta = \arccos\left(\frac{4}{9\sqrt{2}}\right) = 71.681...^\circ$. 6. State the final angle: $\mathbf{71.7^\circ}$.

Line $L_1$ has vector equation $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and line $L_2$ has equation $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + s \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$. Show analytically that $L_1$ and $L_2$ are skew lines.

1. Check if the lines are parallel. Since the direction vector $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ is not a scalar multiple of $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$, the lines are not parallel. 2. Equate the $x$ and $y$ parametric components of the lines to test for intersection points: $1 + t = 2 + 2s \implies t - 2s = 1$ (Equation 1), and $2 - t = 1 + s \implies -t - s = -1 \implies t + s = 1$ (Equation 2). 3. Solve this simultaneous system. Subtracting Eq 2 from Eq 1 gives $-3s = 0 \implies s = 0$. Substituting $s=0$ into Eq 2 yields $t = 1$. 4. Substitute these parameter values into the $z$-components of each respective line to see if they match. For $L_1$ ($t=1$): $z = 3 + (1) = 4$. 5. Substitute for $L_2$ ($s=0$): $z = 0 + (-1)(0) = 0$. 6. Conclude : Because the $z$-coordinates do not match ($4 \neq 0$), the lines do not intersect. Since they are neither parallel nor intersecting, they must be skew lines .

Consider the two lines defined by: \[L_1: \mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \quad \text{and} \quad L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\] Show algebraically that $L_1$ and $L_2$ intersect, and find the coordinates of their point of intersection.

Equate the components of $L_1$ and $L_2$: M1 $2 + \lambda = 3 + \mu$ (1), $2\lambda = 2 - \mu$ (2), $-1 + \lambda = \mu$ (3) A1 Substitute (3) into (2): $2\lambda = 2 - (\lambda - 1) = 3 - \lambda$ M1 $\lambda = 1$, $\mu = 0$ A1 Check in (1): $2 + 1 = 3$ and $3 + 0 = 3$ ✓, so all three equations are satisfied and the lines intersect R1 Substitute $\lambda = 1$ into $L_1$: point of intersection $(3, 2, 0)$ A1

Home › IB Maths AA HL › Questions by topic › Vector Lines

IB Maths AA HL · Unit 3: Geometry and Trigonometry

IB Maths AA HL Vector Lines Questions

Updated October 2026

Exam-style IB Maths AA HL vector lines questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step, showing the working that earns the marks — then try the practice questions and check your working against the mark scheme.

53 questions
Paper 1: 38
Paper 2: 15
10 easy
21 medium
21 hard
1 very hard
3 worked examples

Practise Vector Lines questions →

🖩 How do I do this on my calculator? Keystrokes for the Casio fx-CG50, TI-84 Plus CE and TI-Nspire.

What you need to know

The cross product a × b gives a vector perpendicular to both. HL AA Paper 1 tests the coordinate formula and the |a × b| = |a||b|sinθ identity. 3D vectors — dot product, cross product, and applications overview →

Show that the diagonals of a parallelogram bisect each other, or that three points are collinear. HL AA Paper 1 tests vector proofs every year. Vector proofs and geometric arguments overview →

What's examined in AA HL vector lines

The question bank covers these vector lines question types (number of questions in brackets):

Vector Line Equations & Intersection (19)
Vector Product Applications (17)
Line Distances & Angles (17)

Key formulas

Vector magnitude: $| \vec{a} | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$
Vector equation of a line: $\vec{r} = \vec{a} + t \vec{d}$
Dot product: $\vec{a} \cdot \vec{b} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = | \vec{a} | | \vec{b} | \cos θ$

In the same notation as the IB formula booklet. All AA HL formulas →

Explained, with a worked example: Magnitude of a vector · Scalar (dot) product

Where students lose marks on vector lines

Examiner's noteVectors: skew lines without the non-parallel check, and planes

Where marks are lost: November 2025: many did not show that the lines were not parallel before concluding they were skew, made errors solving the simultaneous equations (often with scattered working), and did not give a plane's equation with integer coefficients. May 2024: the relationship between planes was often misread; the report recommends diagrams.

Examiners flagged this in 2 exam series (May 2024, November 2025). Summarised in our own words from IB subject reports, IB Programme Resource Centre. Read the full note →

Vector Lines worked examples

Worked example 1: The vector product and area · easy

Given the vectors $a = i + 2 j - k$ and $b = 3 i + k$ , find the exact area of the parallelogram formed by these two vectors.

Solution

6 steps · 1 to fill in

Examiner tip: A common mistake is forgetting the negative sign on the middle (

j

) component when calculating the cross product manually via the determinant method.

Worked example 2: Angle between two vector lines · medium

Find the acute angle between the line $L_{1} : r = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) + t (\begin{matrix} 2 \\ - 1 \\ 2 \end{matrix})$ and the line $L_{2} : r = (\begin{matrix} 0 \\ - 2 \\ 3 \end{matrix}) + s (\begin{matrix} 1 \\ 4 \\ - 1 \end{matrix})$ . Give your answer in degrees to one decimal place.

Solution

6 steps · 1 to fill in

Examiner tip: Only use the direction vectors (the multipliers of

t

and

s

) to find the angle between lines; the position vectors (starting points) are completely irrelevant to the angle and plugging them into the formula is a frequent conceptual error.

Worked example 3: Proving lines are skew · hard

Line $L_{1}$ has vector equation $r = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) + t (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$ and line $L_{2}$ has equation $r = (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}) + s (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})$ . Show analytically that $L_{1}$ and $L_{2}$ are skew lines.

Solution

6 steps · 1 to fill in

Examiner tip: You must explicitly state and verify that the direction vectors are not scalar multiples of each other before testing for intersection; otherwise, you haven't fully proven they are skew rather than just coincident or parallel.

Try these IB Maths AA HL vector lines questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 3 marks · Paper 1

Find the vector product $a \times b$ where $a = (\begin{matrix} 1 \\ - 2 \\ 3 \end{matrix})$ and $b = (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})$ .

Attempt it and see the mark scheme →

Question 2 · medium · 6 marks · Paper 1

Consider the two lines defined by: $L_{1} : r = (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}) + λ (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) and L_{2} : r = (\begin{matrix} 3 \\ 2 \\ 0 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$ Show algebraically that $L_{1}$ and $L_{2}$ intersect, and find the coordinates of their point of intersection.

Full worked solution Free, no sign-in

Equate the components of $L_{1}$ and $L_{2}$ : M1
$2 + λ = 3 + μ$ (1), $2 λ = 2 - μ$ (2), $- 1 + λ = μ$ (3) A1
Substitute (3) into (2): $2 λ = 2 - (λ - 1) = 3 - λ$ M1
$λ = 1$ , $μ = 0$ A1
Check in (1): $2 + 1 = 3$ and $3 + 0 = 3$ ✓, so all three equations are satisfied and the lines intersect R1
Substitute $λ = 1$ into $L_{1}$ : point of intersection $(3, 2, 0)$ A1

Every question here has a full worked solution. See them all →

Question 3 · hard · 7 marks · Paper 2

Two lines

L_{1}

and

L_{2}

are given by the vector equations:

L_{1} : r = (\begin{matrix} 1 \\ 0 \\ 2 \end{matrix}) + λ (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})

L_{2} : r = (\begin{matrix} - 1 \\ 2 \\ 1 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 3 \end{matrix})

(a) Show that the lines $L_{1}$ and $L_{2}$ are skew.

(b) Find the shortest distance between the lines $L_{1}$ and $L_{2}$ . Give your answer in the form $\frac{a}{\sqrt{b}}$ where $a, b \in Z^{+}$ .

Attempt it and see the mark scheme →

Free account — sign in to download Topic PDF: 7 questions with worked answers · 4 pages · 205 KB

All 53 vector lines questions with mark schemes →

Save your progress and get 5 more questions on every topic, checked instantly. Free account, no card. Join free

Revise vector lines for the exam

FAQ

How many IB Maths AA HL vector lines questions are there?

There are 53 exam-style vector lines questions in the AA HL question bank (Paper 1: 38 · Paper 2: 15), graded 10 easy, 21 medium, 21 hard, 1 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is vector lines on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 38 · Paper 2: 15. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA HL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

What else is free here

Save your answers and track your progressA free account remembers the questions you have done on every topic and brings you back to this page.

Join free

Already have an account? Sign in

IB Maths AA HL Vector Lines Questions

Try it first

What you need to know

What's examined in AA HL vector lines

Key formulas

Where students lose marks on vector lines

Vector Lines worked examples

Worked example 1: The vector product and area · easy

Solution

Worked example 2: Angle between two vector lines · medium

Solution

Worked example 3: Proving lines are skew · hard

Solution

Try these IB Maths AA HL vector lines questions

Question 1 · easy · 3 marks · Paper 1

Question 2 · medium · 6 marks · Paper 1

Full worked solution Free, no sign-in

Question 3 · hard · 7 marks · Paper 2

Revise vector lines for the exam

FAQ

What else is free here

More AA HL Unit 3 topics